MODULAR FORMS-page49

MODULAR FORMS-page49 - M = ( 1 2 0 1 ). Then the assertion...

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45 Proof. We shall prove in the next lecture that it is enough to check this for generators of the group SL(2 , Z ). Also we shall show that the group SL(2 , Z ) is generated by the matrices M 1 = ( 1 1 0 1 ) ,M 2 = ( 0 - 1 1 0 ) , - I. We have from (4.14) and (4.15) f ( τ + 1) 8 = f ( τ ) 8 , f ( - 1 ) 8 = τ 12 f ( τ ) 8 . This proves the assertion. Corollary 5.3. Let η ( τ ) be the Dedekind η -function. Then η ( τ ) 24 = q Y m =1 (1 - q m ) 24 , q = e 2 πiτ satisfies η ( ατ + β γτ + δ ) 24 = ( γτ + δ ) 12 η ( τ ) 24 . Proof. Use (4.10) ϑ 0 1 2 1 2 = - 2 πη ( τ ) 3 . Corollary 5.4. Let M = α β γ δ « SL(2 , Z ) . Assume that the products αβ,γδ are even. Then Θ( z γτ + δ ; ατ + β γτ + δ ) = ζ ( γτ + δ ) 1 2 e πiγz 2 / ( γτ + δ ) Θ( z ; τ ) , (5.18) where ζ 8 = 1 and the branch of the square root is chosen to have non-negative real part. Proof. Recall that Θ( z ; τ ) = ϑ 00 ( z ; τ ), so Theorem 5.1 gives immediately that Θ( z γτ + δ ; ατ + β γτ + δ ) = c ( M,τ ) e πiγz 2 / ( γτ + δ ) Θ( z ; τ ) for some constant c ( M,τ ) depending only on M and τ . Take M = ` 0 1 - 1 0 ´ . Then formula (5.13) checks the assertion in this case. Take
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Unformatted text preview: M = ( 1 2 0 1 ). Then the assertion follows from (5.10). Now we argue by induction on | | + | | . If | | > | | , using that ( z, + 2) = ( z ; ), we substitute 2 in (5.16) to obtain that the assertion is true for M = 2 2 . Since we can decrease | 2 | in this way, the assertion will follow by induction. Note that we used that | 2 | is not equal to | | or | | because ( , ) = 1 and is even. Now, if | | < | | , we use the substitution -1 / . Using (5.13) we see that the asssertion for M follows from the assertion fo M = - - . This reduces again to the case | | > | | ....
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This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.

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